Chapter 5: Q84CE (page 193)

Determine the expectation value of the momentum of the particle. Explain.

Short Answer

Expert verified

The expected value of momentum is $i ℏ (\frac{1 - a^{2}}{a})$ .

Step by step solution

Define expectation value

A particle of mass is represented by a wavefunction,

$ψ (x) = {\begin{cases} 2 \sqrt{a^{3}} x e^{- a x} & x > 0 \\ 0 & x < 0 \end{cases}}$

The expectation value of $<\overset{⏞}{A}>$ is defined as

$<\hat{A}> = \int_{- \infty}^{\infty} ψ^{*} (x) \hat{A} ψ (x)$

Determine the expectation value of momentum.

The expression for momentum operator is

$\hat{p} = - i ℏ \frac{\partial}{\partial x}$

The expectation value of momentum is

$\begin{array}{rcl} <\hat{p}> & = & \int_{- \infty}^{\infty} ψ^{*} (x) \hat{p} ψ (x) \\ = & - i ℏ \int_{0}^{\infty} (2 \sqrt{a^{3}} x e^{- a x}) \frac{\partial}{\partial x} (2 \sqrt{a^{3}} x e^{- a x}) \\ = & - i ℏ \int_{0}^{\infty} (2 \sqrt{a^{3}} x e^{- a x}) [2 \sqrt{a^{3}} (e^{- a x} + x \frac{e^{- a x}}{- a})] \\ = & \frac{i ℏ}{a} \int_{0}^{\infty} (2 \sqrt{a^{3}} x e^{- a x}) (2 \sqrt{a^{3}} x e^{- a x}) - i ℏ \int_{0}^{\infty} (2 \sqrt{a^{3}} x e^{- a x}) (2 \sqrt{a^{3}} e^{- a x}) \end{array}$

Solve further as:

$<\hat{p}> = \frac{i ℏ}{a} \int_{0}^{\infty} ψ^{2} (x) - 4 a^{3} i ℏ \int_{0}^{\infty} (x e^{- 2 a x})$

The integral in the first term will be equal to one as $ψ (x)$ is normalised. Solving integral in the second term separately as follows:

$\begin{array}{rcl} l & = & \int_{0}^{\infty} (x e^{- 2 a x}) \\ = & x \frac{e^{- 2 a x}}{- 2 a} - \int_{}^{} (\frac{d x}{d x}) (\int_{}^{} e^{- 2 a x}) \\ = & {- \frac{x e^{- 2 a x}}{2 a} - \frac{e^{- 2 a x}}{4 a^{2}}|}_{0}^{\infty} \\ = & \frac{1}{4 a^{2}} \end{array}$

Inserting this result in the above expression for momentum.

$\begin{array}{rcl} <\hat{p}> & = & \frac{i ℏ}{a} - 4 a^{3} i ℏ (\frac{1}{4 a^{2}}) \\ = & i ℏ (\frac{1}{a} - a) \\ = & i ℏ (\frac{1 - a^{2}}{a}) \end{array}$